Question

Suppose $y=f\left(x\right)$ and $y=g\left(x\right)$ are two functions whose graphs intersect at the three points $(0,4),(2,2)$ and $(4,0)$. And also $f\left(x\right)>g\left(x\right)$ for $x\in (0,2),f\left(x\right)<g\left(x\right)$ for $x\in (2,4)$. If $\underset{0}{\overset{4}{\int }}\left(f\right(x)-g(x\left)\right)dx=10$ and $\underset{2}{\overset{4}{\int }}\left(g\right(x)-f(x\left)\right)dx=5$, then the area between the two curves for $x\in (0,2)$ is

• A. $20$
• B. $15$
• C. $10$
• D. $5$

Answer 1

The correct option is B $15$

Given: $y=f\left(x\right)$ and $y=g\left(x\right)$ intersect at $(0,4),(2,2),(4,0)$
$f\left(x\right)>g\left(x\right)$ for $x\in (0,2)$
$f\left(x\right)<g\left(x\right)$ for $x\in (2,4)$

Inferences: $f\left(0\right)=g\left(0\right)=4;f\left(2\right)=g\left(2\right)=2$
$f\left(4\right)=g\left(4\right)=0$

Area $=\underset{0}{\overset{2}{\int }}\left(f\right(x)-g(x\left)\right)dx$

Given $\underset{0}{\overset{4}{\int }}\left(f\right(x)-g(x\left)\right)dx=10$
$\Rightarrow \underset{0}{\overset{2}{\int }}\left(f\right(x)-g(x\left)\right)dx+\underset{2}{\overset{4}{\int }}\left(f\right(x)-g(x\left)\right)dx=10$
$\Rightarrow \underset{0}{\overset{2}{\int }}\left(f\right(x)-g(x\left)\right)dx-\underset{2}{\overset{4}{\int }}\left(g\right(x)-f(x\left)\right)dx=10$
$\Rightarrow \underset{0}{\overset{2}{\int }}\left(f\right(x)-g(x\left)\right)dx=10+5=15\text{sq. units}$